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جستجو

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With respect to the origin $O$, the position vectors of two points $A$ and $B$ are given by $\overrightarrow {OA}  = i + 2j + 2k$ and $\overrightarrow {OB}  = 3i + 4j$. The point $P$ lies on the line through $A$ and $B$, and $\overrightarrow {AP}  = \lambda \overrightarrow {AB} $.

a) Show that $\overrightarrow {OP}  = \left( {1 + 2\lambda } \right)i + \left( {2 + 2\lambda } \right)j + \left( {2 - 2\lambda } \right)k$.

b) By equating expressions for $\cos AOP$ and $\cos BOP$ in terms of in terms of $\lambda $, find the value of $\lambda $ for which $OP$ bisects the angle $AOB$.

c) When $\lambda $ has this value, verify that $AP:PB = OA:OB$.

پاسخ تشریحی :
نمایش پاسخ

a) Use a correct method to express $\overrightarrow {OP} $ in terms of $\lambda $

Obtain the given answer

b) EITHER: Use correct method to express scalar product of $\overrightarrow {OA} $ and $\overrightarrow {OP} $, or $\overrightarrow {OB} $ and $\overrightarrow {OP} $ in terms of $\lambda $

Using the correct method for the moduli, divide scalar products by products of moduli and express $\cos AOP = \cos BOP$ in terms of $\lambda $, or in terms of $\lambda $ and $OP$

OR1: Use correct method to express $O{A^2} + O{P^2} - A{P^2}$, or $O{B^2} + O{P^2} - B{P^2}$ in terms of $\lambda $

Using the correct method for the moduli, divide each expression by twice the product of the relevant moduli and express $\cos AOP = \cos BOP$ in terms of $\lambda $, or $\lambda $ and $OP$

Obtain a correct equation in any form, e.g. $\frac{{9 + 2\lambda }}{{3\sqrt {\left( {9 + 4\lambda  + 12{\lambda ^2}} \right)} }} = \frac{{11 + 14\lambda }}{{5\sqrt {\left( {9 + 4\lambda  + 12{\lambda ^2}} \right)} }}$

Solve for $\lambda $

Obtain $\lambda  = \frac{3}{8}$

c) Verify the given statement correctly

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